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I have the following problem. I need to find the volume of the shape that is bounded up by $$u(x,y)=13+x^2$$ and down by $$v(x,y)=7x+4y$$ and from its sides by the cylinder $$x^2+y^2=1.$$ Now I know I need to do to the integral $\displaystyle \iint\limits_D \left(u(x,y)-v(x,y) \right)dxdy$

But I dont really know how to define $D$, I know that is by the projection in $XY$ plane.

I tried to draw it in the $XY$ plane, drew the cylinder and the plane $z=7x+4y$. But I don't really know where to go next.

Some tips will really help me! :)

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hmm no one answered, so I guess that everybody didnt understood bad (english not my mothertongue)

anyway. the projection is the cylinder, and after moving to polar cordinates the integral is easy. I just had a calculation error that made me think it was more complicated... answer is $13\pi + \pi/4 $ if anyone tried.

thank you anyway!!

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Neither $u(x,y)$ nor $v(x,y)$ define a curve, which would usually be the boundary of a region. You should have something like $u(x,y)=k$ for some given $k$. Then there is no sign of $z$ in the problem except when you talk about volume and a cylinder. As there is no limitation on $z$, the volume is infinite. Maybe what you really want is $\displaystyle \iint\limits_D \left(u(x,y)-v(x,y) \right)dxdy$ with $D$ being the region inside the circle $x^2+y^2=1?$ That is at least a problem that can be solved. You would set it up as $$\int_{-1}^1 \int _{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}13+x^2-7x-4y\; dy\; dx$$

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Another way of doing this is triple integrating with cylindrical coordinates:$$\\x=r\cos\theta\\y=r\sin\theta\\z=z$$ Volume is given by:
$$\iiint_D \,dV$$ Since the cylinder is giving your bound for $r$, $r\in[0, 1]$. $\theta$, as usual, is $\theta\in[0, 2π]$. The z bounds is the hardest part. However, you can pretty easily see that $u$ is always greater than $v$, therefore $z\in[7x+4y,13+x^2]$, or in cylindrical coordinates $z\in[7r\cos\theta+4r\sin\theta, r^2\cos^2\theta+13]$
Our integral becomes: $$V=\int_0^{2\pi}\int_0^1\int_{7r\cos\theta+4r\sin\theta}^{r^2\cos^2\theta+13}r\,dz\,dr\,d\theta.$$
This isn't the most fun integral to do, but it comes out to $\frac{53\pi}{4}.$ $\blacksquare$

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